Chemistry
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Experiment-8SingleReplacement-Macmillan1.docxExperiment-8: Single Replacement Reactions – Macmillan
1) The metal LOSES electrons and DISSOLVES in the solution to form cations. It is OXIDIZED.
2) The metal ions in solution GAIN elections and are converted to a SOLID. They are REDUCED.
We’re looking at reactions in water, so one of the reactants starts out as a homogeneous, aqueous solution (aq). The other reactant is a solid metal, which we indicate by writing (s) after the symbol. Note: (s) means “solid”, NOT “soluble.” If a reaction occurs, the first sign will be a color change on the surface of the metal. Most likely, it will darken as it starts to dissolve in the solution. The solution color may change as well, as transition metal ions are often colored.
2AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + 2Ag(s)
In this example, a colorless solution (silver chloride) reacts with copper metal, and a solid silver precipitate forms. Copper dissolves in the solution as copper(II) ions, which are blue in water. Remember that when ionic compounds dissolve in water, their ions separate (Complete Ionic Equation):
2Ag(aq) + 2NO3(aq) + Cu(s) 2Ag(s) + Cu+2(aq) + 2NO3-2(aq)
Note that the nitrate ions didn’t do anything in this reaction. We can write the Net Ionic Equation, leaving out these so-called “spectator ions:”
2Ag(aq) + Cu(s) 2Ag(s) + Cu+2(aq)
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Oxidant |
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Reductant |
E° (V) |
We can predict whether any metal will react with any other metal ion by comparing their reduction potentials, as shown is this table. If the metal is higher on the table (that is, if it has the more NEGATIVE reduction potential) then it can “displace” the other metal ion in solution, as described above. The bigger the potential difference, the more energy that is released by the reaction. If we connect wires to two different metals and place them in an electrolyte solution, the voltage we measure should be the difference between the reduction potential of the metal that is oxidized the reduction potential of the cation that is reduced. |
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Mg2+(aq) + 2 e− |
⇌ |
Mg(s) |
−2.372 |
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Al3+(aq) + 3 e− |
⇌ |
Al(s) |
−1.662 |
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Mn2+(aq) + 2 e− |
⇌ |
Mn(s) |
−1.185 |
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Zn2+(aq) + 2 e− |
⇌ |
Zn(s) |
−0.762 |
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Fe2+(aq) + 2 e− |
⇌ |
Fe(s) |
−0.440 |
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Co2+(aq) + 2 e− |
⇌ |
Co(s) |
-0.280 |
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Pb2+(aq) + 2 e− |
⇌ |
Pb(s) |
−0.126 |
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2H+(aq) + e− |
⇌ |
H2(g) |
0.000 |
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Cu2+(aq) + 2 e− |
⇌ |
Cu(s) |
+0.337 |
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Ag+(aq) + e− |
⇌ |
Ag(s) |
+0.800 |
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As you can see, the entire reduction potential scale is referenced to the H+/H2 potential, which is DEFINED as ZERO VOLTS. Any metal that is higher on the reduction scale than hydrogen – that is, any metal with a NEGATIVE reduction potential – can reduce H+ to hydrogen gas. We call such metals “Active Metals.”
Your Mission: Confirm the Metal Activity Series
Chapter 8.7 includes a simulation of these reactions (Figure 8.24), in which you can select a metal from the left column (shown in the left hand column of the table below) and a metal ion compound from the right-hand column (shown across the top of the table below). Here’s the link:
https://savi-cdn.macmillanlearning.com/revell-videopicker-activityseries/index.html
For each reaction, record your observation: Yes (reaction dissolves the metal) or No (no reaction)
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Mg(NO3)2(aq) |
Zn(NO3)2(aq) |
Fe(NO3)3(aq) |
Cu(NO3)2(aq) |
AgNO3(aq) |
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Mg(s) |
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Zn(s) |
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Fe(s) |
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Cu(s) |
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Ag(s) |
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At least one of these reactions creates a gas. Which reaction(s)? What is the gas? What does this observation suggest about the ionic solution used in the reaction(s)? Write the Net Ionic Equation for the gas-forming reaction:
For each combination that you mark “Yes” write the Net Ionic Equation:
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